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Oliga [24]
3 years ago
13

Find the reference angle of 100 degrees

Mathematics
1 answer:
Roman55 [17]3 years ago
4 0

Since the angle  

100 °  is in the second quadrant, the reference angle formula is  A r = 180 ° −

A c.   A r = 180 ° − 100 °

The reference angle is  A r = 80 ° .

80 °

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Tyra buys 14 gallons of gas for $35. How could she figure out how much 1 gallon of gas cost
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Answer:

The cost of 1 gallon is $2.5

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Step-by-step explanation:

Given

Cost of 14 gallons = $35

Required

Determine the cost of 1 gallon

From the given parameters, we have:

14\ gallons = \$35

Divide both sides by 14

\frac{14\ gallons}{14} = \frac{\$35}{14}

1\ gallon = \frac{\$35}{14}

1\ gallon = \$2.5

Hence,

<em>The cost of 1 gallon is $2.5</em>

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Which expression represents the product of b and 18?
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Answer:

You're answer is D.18b

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The horizontal line segment at the top of the polygon on the grid below is how many units long?​
Ymorist [56]

Answer:

6 units long.

Step-by-step explanation:

Given:

A polygon drawn on a graph.

In order to determine the distance of the horizontal line at the top, we find the coordinates of the end points and then use distance formula to find the exact distance.

Let us label the polygon as ABCD as shown below. AB is the length of the horizontal line.

Coordinates of A are (x_1,y_1)=(-3, 4) as seen in the graph.

Coordinates of B are (x_2,y_2)=( 3, 4) as seen in the graph.

Now, distance formula  for two points (x_1,y_1)\ and\ (x_2,y_2) is:

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\d_{AB}=\sqrt{(3-(-3))^2+(4-4)^2}\\d_{AB}=\sqrt{(3+3)^2+0}\\d_{AB}=\sqrt{36}=6\ units

Therefore, the horizontal line at the top is 6 units long.

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3 years ago
A married man filing serpartly with a taxable income of 167,000
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Let X denote the amount of time a book on two-hour reserve is actually checked out, and suppose the cdf is the following. F(x) =
Troyanec [42]

Answer:

a) P (x <= 3 ) = 0.36

b) P ( 2.5 <= x <= 3  ) = 0.11

c) P (x > 3.5 ) = 1 - 0.49 = 0.51

d) x = 3.5355

e) f(x) = x / 12.5

f) E(X) = 3.3333

g) Var (X) = 13.8891  , s.d (X) = 3.7268

h) E[h(X)] = 2500

Step-by-step explanation:

Given:

The cdf is as follows:

                           F(x) = 0                  x < 0

                           F(x) = (x^2 / 25)     0 < x < 5

                           F(x) = 1                   x > 5

Find:

(a) Calculate P(X ≤ 3).

(b) Calculate P(2.5 ≤ X ≤ 3).

(c) Calculate P(X > 3.5).

(d) What is the median checkout duration ? [solve 0.5 = F()].

(e) Obtain the density function f(x). f(x) = F '(x) =

(f) Calculate E(X).

(g) Calculate V(X) and σx. V(X) = σx =

(h) If the borrower is charged an amount h(X) = X2 when checkout duration is X, compute the expected charge E[h(X)].

Solution:

a) Evaluate the cdf given with the limits 0 < x < 3.

So, P (x <= 3 ) = (x^2 / 25) | 0 to 3

     P (x <= 3 ) = (3^2 / 25)  - 0

     P (x <= 3 ) = 0.36

b) Evaluate the cdf given with the limits 2.5 < x < 3.

So, P ( 2.5 <= x <= 3 ) = (x^2 / 25) | 2.5 to 3

     P ( 2.5 <= x <= 3  ) = (3^2 / 25)  - (2.5^2 / 25)

     P ( 2.5 <= x <= 3  ) = 0.36 - 0.25 = 0.11

c) Evaluate the cdf given with the limits x > 3.5

So, P (x > 3.5 ) = 1 - P (x <= 3.5 )

     P (x > 3.5 ) = 1 - (3.5^2 / 25)  - 0

     P (x > 3.5 ) = 1 - 0.49 = 0.51

d) The median checkout for the duration that is 50% of the probability:

So, P( x < a ) = 0.5

      (x^2 / 25) = 0.5

       x^2 = 12.5

      x = 3.5355

e) The probability density function can be evaluated by taking the derivative of the cdf as follows:

       pdf f(x) = d(F(x)) / dx = x / 12.5

f) The expected value of X can be evaluated by the following formula from limits - ∞ to +∞:

         E(X) = integral ( x . f(x)).dx          limits: - ∞ to +∞

         E(X) = integral ( x^2 / 12.5)    

         E(X) = x^3 / 37.5                    limits: 0 to 5

         E(X) = 5^3 / 37.5 = 3.3333

g) The variance of X can be evaluated by the following formula from limits - ∞ to +∞:

         Var(X) = integral ( x^2 . f(x)).dx - (E(X))^2          limits: - ∞ to +∞

         Var(X) = integral ( x^3 / 12.5).dx - (E(X))^2    

         Var(X) = x^4 / 50 | - (3.3333)^2                         limits: 0 to 5

         Var(X) = 5^4 / 50 - (3.3333)^2 = 13.8891

         s.d(X) = sqrt (Var(X)) = sqrt (13.8891) = 3.7268

h) Find the expected charge E[h(X)] , where h(X) is given by:

          h(x) = (f(x))^2 = x^2 / 156.25

  The expected value of h(X) can be evaluated by the following formula from limits - ∞ to +∞:

         E(h(X))) = integral ( x . h(x) ).dx          limits: - ∞ to +∞

         E(h(X))) = integral ( x^3 / 156.25)    

         E(h(X))) = x^4 / 156.25                       limits: 0 to 25

         E(h(X))) = 25^4 / 156.25 = 2500

8 0
3 years ago
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